Write an equation for an ellipse centered at the origin, which has foci at $(0,\pm3)$ and co-vertices at $(\pm2,0)$.
Explanation: The strategy In order to find the equation of the ellipse, we perform the following steps. Find the focal length, $f$, and the minor radius, $q$, using the given information about the center, the foci, and the co-vertices. Find the major radius, $p$, using the equation $f^2=p^2-q^2$. Match the major and minor radii to the vertical and horizontal radii by determining the axis along which the foci lie. Substitute all the values we found in the standard equation of an ellipse. Finding the focal length and minor radius The focal length is $3$ units, since the foci are at $(0,\pm 3)$. [How did we know that?] The minor radius is $2$ units, since the co-vertices are at $(\pm 2,0)$. [How did we know that?] Finding the major radius We've found that the focal length, $f$, is $3$ units, and the minor radius, $q$, is $2$ units. Let's substitute these values into the equation to find the major radius, $p$. $\begin{aligned}f^2&=p^2-q^2\\\\ 3^2&=p^2-2^2\\\\ 9+4&=p^2\\\\ \sqrt{13}&=p\end{aligned}$ Therefore, the major radius is $\sqrt{13}$ units. Matching the major and minor radii with the vertical and horizontal radii Since the foci are located on the $y$ -axis, the major radius is the vertical radius. [Why?] In consequence, the minor radius is the horizontal radius. This means that the vertical radius is $\sqrt{13}$ units and the horizontal radius is $2$ units. Writing the equation Our ellipse is centered at $(C 0, 0)$, has a horizontal radius of $ {2}$ units and a vertical radius of $ {\sqrt{13}}$ units. So it can be represented by the equation below. [How did we get this equation?] $\begin{aligned}\dfrac{(x - C 0)^2}{ 2^2} + \dfrac{(y - 0)^2}{( {\sqrt{13}})^2} &= 1\\\\\\ \dfrac{x^2}{ {4}} + \dfrac{y^2}{ {13}} &= 1 &(\text{Simplify terms})\end{aligned}$ [Can we write the equation in another form?] Summary The ellipse in question can be represented by the following equation. $\dfrac{x^2}{4} + \dfrac{y^2}{ 13} = 1$